ScienceAges 10-12

Series & parallel resistor networks

Calculate equivalent resistance for resistors in series (R₁ + R₂) or in parallel ((R₁ × R₂) ÷ (R₁ + R₂)). Review suitability for your learners before teaching.

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Age band:Ages 10-12 Ages 13-15

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Worksheet StudioPractice pack · Ages 10-12

Series & parallel resistor networks

Take your time. Show your thinking.

1. Two resistors of 24 Ω and 48 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 24 Ω ∥ 48 Ω · 1/R_eq = 1/24 + 1/48
2. Two resistors of 6 Ω and 30 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 6 Ω ∥ 30 Ω · 1/R_eq = 1/6 + 1/30
3. Two resistors of 6 Ω and 12 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 6 Ω ∥ 12 Ω · 1/R_eq = 1/6 + 1/12
4. Two resistors of 24 Ω and 72 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 24 Ω ∥ 72 Ω · 1/R_eq = 1/24 + 1/72
5. Two resistors of 12 Ω and 24 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 12 Ω ∥ 24 Ω · 1/R_eq = 1/12 + 1/24
6. Two resistors of 24 Ω and 16 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 24 Ω, R₂ = 16 Ω · R_eq = R₁ + R₂ = ?
7. Two resistors of 12 Ω and 36 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 12 Ω ∥ 36 Ω · 1/R_eq = 1/12 + 1/36
8. Two resistors of 24 Ω and 120 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 24 Ω ∥ 120 Ω · 1/R_eq = 1/24 + 1/120
9. Two resistors of 22 Ω and 8 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 22 Ω, R₂ = 8 Ω · R_eq = R₁ + R₂ = ?
10. Two resistors of 7 Ω and 15 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 7 Ω, R₂ = 15 Ω · R_eq = R₁ + R₂ = ?
11. Two resistors of 21 Ω and 18 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 21 Ω, R₂ = 18 Ω · R_eq = R₁ + R₂ = ?
12. Two resistors of 16 Ω and 48 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 16 Ω ∥ 48 Ω · 1/R_eq = 1/16 + 1/48

12 questions · Seed 51 · Age guidance, not curriculum certification

View the answer key

Worksheet StudioAnswer key · Ages 10-12

Series & parallel resistor networks

Answers and reasoning. Review before teaching.

1. Two resistors of 24 Ω and 48 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
16 Ω
For two parallel resistors, R_eq = (24 × 48) ÷ (24 + 48) = 1152 ÷ 72 = 16 Ω.
2. Two resistors of 6 Ω and 30 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
5 Ω
For two parallel resistors, R_eq = (6 × 30) ÷ (6 + 30) = 180 ÷ 36 = 5 Ω.
3. Two resistors of 6 Ω and 12 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
4 Ω
For two parallel resistors, R_eq = (6 × 12) ÷ (6 + 12) = 72 ÷ 18 = 4 Ω.
4. Two resistors of 24 Ω and 72 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
18 Ω
For two parallel resistors, R_eq = (24 × 72) ÷ (24 + 72) = 1728 ÷ 96 = 18 Ω.
5. Two resistors of 12 Ω and 24 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
8 Ω
For two parallel resistors, R_eq = (12 × 24) ÷ (12 + 24) = 288 ÷ 36 = 8 Ω.
6. Two resistors of 24 Ω and 16 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
40 Ω
In a series circuit, resistances add directly: R_eq = 24 Ω + 16 Ω = 40 Ω.
7. Two resistors of 12 Ω and 36 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
9 Ω
For two parallel resistors, R_eq = (12 × 36) ÷ (12 + 36) = 432 ÷ 48 = 9 Ω.
8. Two resistors of 24 Ω and 120 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
20 Ω
For two parallel resistors, R_eq = (24 × 120) ÷ (24 + 120) = 2880 ÷ 144 = 20 Ω.
9. Two resistors of 22 Ω and 8 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
30 Ω
In a series circuit, resistances add directly: R_eq = 22 Ω + 8 Ω = 30 Ω.
10. Two resistors of 7 Ω and 15 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
22 Ω
In a series circuit, resistances add directly: R_eq = 7 Ω + 15 Ω = 22 Ω.
11. Two resistors of 21 Ω and 18 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
39 Ω
In a series circuit, resistances add directly: R_eq = 21 Ω + 18 Ω = 39 Ω.
12. Two resistors of 16 Ω and 48 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
12 Ω
For two parallel resistors, R_eq = (16 × 48) ÷ (16 + 48) = 768 ÷ 64 = 12 Ω.

12 questions · Seed 51 · Age guidance, not curriculum certification

Worksheet StudioPractice pack · Ages 10-12

Series & parallel resistor networks

Take your time. Show your thinking.

1. Two resistors of 24 Ω and 48 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 24 Ω ∥ 48 Ω · 1/R_eq = 1/24 + 1/48
2. Two resistors of 6 Ω and 30 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 6 Ω ∥ 30 Ω · 1/R_eq = 1/6 + 1/30
3. Two resistors of 6 Ω and 12 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 6 Ω ∥ 12 Ω · 1/R_eq = 1/6 + 1/12
4. Two resistors of 24 Ω and 72 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 24 Ω ∥ 72 Ω · 1/R_eq = 1/24 + 1/72
5. Two resistors of 12 Ω and 24 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 12 Ω ∥ 24 Ω · 1/R_eq = 1/12 + 1/24
6. Two resistors of 24 Ω and 16 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 24 Ω, R₂ = 16 Ω · R_eq = R₁ + R₂ = ?
7. Two resistors of 12 Ω and 36 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 12 Ω ∥ 36 Ω · 1/R_eq = 1/12 + 1/36
8. Two resistors of 24 Ω and 120 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 24 Ω ∥ 120 Ω · 1/R_eq = 1/24 + 1/120
9. Two resistors of 22 Ω and 8 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 22 Ω, R₂ = 8 Ω · R_eq = R₁ + R₂ = ?
10. Two resistors of 7 Ω and 15 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 7 Ω, R₂ = 15 Ω · R_eq = R₁ + R₂ = ?
11. Two resistors of 21 Ω and 18 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 21 Ω, R₂ = 18 Ω · R_eq = R₁ + R₂ = ?
12. Two resistors of 16 Ω and 48 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 16 Ω ∥ 48 Ω · 1/R_eq = 1/16 + 1/48

12 questions · Seed 51 · Age guidance, not curriculum certification

Worksheet StudioAnswer key · Ages 10-12

Series & parallel resistor networks

Answers and reasoning. Review before teaching.

1. Two resistors of 24 Ω and 48 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
16 Ω
For two parallel resistors, R_eq = (24 × 48) ÷ (24 + 48) = 1152 ÷ 72 = 16 Ω.
2. Two resistors of 6 Ω and 30 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
5 Ω
For two parallel resistors, R_eq = (6 × 30) ÷ (6 + 30) = 180 ÷ 36 = 5 Ω.
3. Two resistors of 6 Ω and 12 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
4 Ω
For two parallel resistors, R_eq = (6 × 12) ÷ (6 + 12) = 72 ÷ 18 = 4 Ω.
4. Two resistors of 24 Ω and 72 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
18 Ω
For two parallel resistors, R_eq = (24 × 72) ÷ (24 + 72) = 1728 ÷ 96 = 18 Ω.
5. Two resistors of 12 Ω and 24 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
8 Ω
For two parallel resistors, R_eq = (12 × 24) ÷ (12 + 24) = 288 ÷ 36 = 8 Ω.
6. Two resistors of 24 Ω and 16 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
40 Ω
In a series circuit, resistances add directly: R_eq = 24 Ω + 16 Ω = 40 Ω.
7. Two resistors of 12 Ω and 36 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
9 Ω
For two parallel resistors, R_eq = (12 × 36) ÷ (12 + 36) = 432 ÷ 48 = 9 Ω.
8. Two resistors of 24 Ω and 120 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
20 Ω
For two parallel resistors, R_eq = (24 × 120) ÷ (24 + 120) = 2880 ÷ 144 = 20 Ω.
9. Two resistors of 22 Ω and 8 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
30 Ω
In a series circuit, resistances add directly: R_eq = 22 Ω + 8 Ω = 30 Ω.
10. Two resistors of 7 Ω and 15 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
22 Ω
In a series circuit, resistances add directly: R_eq = 7 Ω + 15 Ω = 22 Ω.
11. Two resistors of 21 Ω and 18 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
39 Ω
In a series circuit, resistances add directly: R_eq = 21 Ω + 18 Ω = 39 Ω.
12. Two resistors of 16 Ω and 48 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
12 Ω
For two parallel resistors, R_eq = (16 × 48) ÷ (16 + 48) = 768 ÷ 64 = 12 Ω.

12 questions · Seed 51 · Age guidance, not curriculum certification